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I am writing a summary on a work on Fluid Dynamics that develops irrotational flow states that appear to interact amongst each other according to the equations of Electromagnetism http://arxiv.org/abs/1301.7540

So it begins with Euler Equations of inviscid compressible fluid. Apply some constraints and then find a solution. The solution is a Bessel function:

$$\left.\begin{array}{rcl} \dfrac{\partial \mathbf{u}}{\partial t} \;+\; (\mathbf{u}.\nabla)\mathbf{u}& \; = \; -\: \dfrac{1}{\rho} \nabla P \\\ \rho(\mathbf{x}, t)& \; \ll \; 1 & \end{array}\right\rbrace$$ $$\Rightarrow \xi =\: \psi_o(t)\: R_{mn}(\mathbf{x}) \;\;: \\\ \begin{cases} \mathrm{Re}(\xi) &\overset{\underset{\mathrm{def}}{}}{=}\: \dfrac{\rho}{\rho_o} - 1 \\\ \psi_o &\overset{\underset{\mathrm{def}}{}}{=}\; A \: e^{-i\omega_ot} \\\ \displaystyle R_{mn} &\overset{\underset{\mathrm{def}}{}}{=}\; \int_{0}^{2\pi} e^{-i(m{\theta}'\,-\,n\phi )}j_m(\kappa_r\sigma)\kappa_rR_o \mathbf{d} \phi \end{cases} $$

My goal is to do a step-by-step proof of his derivation and learn somethings about such system. Later I would like to derive step-by-step how two such systems interact with each other, if possible. The article is rather dry on the derivations as it assumes these are rather uninteresting and unremarkable.


Update 1: So far I have found online derivations to the Euler equation and a very attractive derivation of Bessel functions with gorgeous physical insights to it:

http://galileo.phys.virginia.edu/classes/311/notes/fluids1/fluids11/node10.html

http://physics.ucsc.edu/~josh/116C.07/bessel/node1.html

I can't apply the derivation of Bessel directly because it starts from the equation: $\nabla^2\mathbf{u}(x, y, z) = 0$ . I don't know how to relate that to the Euler equation of the form $\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u}.\nabla)\mathbf{u} = - \frac{1}{\rho} \nabla P$. Does someone know how the two relate?


Update 2: Carlo Beenakker pointed out that the target solution ignores the effects of the convective term of Euler equation: $(\mathbf{u} \cdot \nabla)\mathbf{u}$

That relates to the reference article in that the author makes the assumption of "low amplitude", meaning $\mathbf{u} \ll 1$

Carlo Beenakker has also given a full answer I am still studying. I hope it is complete but I would appreciate anyone helping. I should take a couple of days.


Background: I am not a professional mathematician or physicist. I know the proper way to do this would take a couple of semesters and do the proper college courses on differential equations, with much calculus background, which I don't have. As my interest is mostly only on this specific set of equations and I don't have a tutor or teacher to help me I would need some points on what would be the fastest way to finish this such that the math is rigorous.

I hope some of you have any interest for this curious approach too. Thank you for helping.

PS: Don't mind the article talks about Quantum Mechanics. Im not interested in that. (I eliminated references to QM to avoid misundertandings.)

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  • $\begingroup$ Not a comment on the math, but if you really want to reformulate Quantum Mechanics in terms of classical fluid dynamics and want to be taken seriously rather than viewed as a crank then you are first obligated to understand how Quantum Mechanics is currently formulated and used in some detail. $\endgroup$ Commented May 17, 2013 at 16:09
  • $\begingroup$ Not useful at all! Your comment is completely off topic. And not original at all. Really. This is not the place for this discussion. If you want to have it, send an email for Robert Brady. If you just want to bash him, join every one at: scottaaronson.com/blog/?p=1255 (Get cred with Scott Aahoronson while you do your rants.) Or if you want to know why this approach is worth it and probably right: vzn1.wordpress.com/2013/02/20/… $\endgroup$
    – user34091
    Commented May 17, 2013 at 16:26
  • $\begingroup$ @Jeff, if you want to help with the math I will appreciate. $\endgroup$
    – user34091
    Commented May 17, 2013 at 16:27
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    $\begingroup$ @d12 Jeff gave you a friendly and serious comment. There is nothing bad about that. $\endgroup$
    – The User
    Commented May 17, 2013 at 16:55
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    $\begingroup$ As far as I can tell, the Bessel function only gives an approximate solution to the equation, so it is unlikely to pop out of a first-principles attempt at an exact solution. It might be best if you broke your question down into simpler pieces, and asked them at math.stackexchange.com or one of the other sites listed in the FAQ. There are nice free materials on fluid mechanics and differential equations at MIT's OpenCourseware, but they require a serious commitment and investment of time to work well. $\endgroup$
    – S. Carnahan
    Commented May 18, 2013 at 0:20

1 Answer 1

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I'll make an attempt at providing the steps you are seeking to go "from Euler equation to Bessel function".

You start from the Euler equation, describing conservation of momentum,

$$\rho\frac{\partial \vec{u}}{\partial t}+\rho\vec{u}\cdot\nabla\vec{u}=-\nabla p$$

and the continuity equation, describing conservation of mass,

$$\frac{\partial\rho}{\partial t}=-\nabla\cdot(\rho \vec{u})$$

These are nonlinear equations, to make them tractable you'll want to linearize them, both in the velocity $\vec{u}$ and in the deviations $\delta\rho=\rho-\rho_0$ of the density from the uniform density $\rho_0$. This approximation throws away lots of interesting physics (shock waves, turbulence,...), but without it no simple solution exists.

The linearized equations read

$$\rho_0\frac{\partial \vec{u}}{\partial t}=-\nabla p$$

$$\frac{\partial\delta\rho}{\partial t}=-\rho_0\nabla\cdot\vec{u}$$

We may also assume a linear relation $p=p_0+C^2\delta\rho$ between the pressure $p$ and the density variations. (This is a socalled adiabatic equation of state, the coefficient $C^2$ must be positive for mechanical stability.) We define $\xi=\delta\rho/\rho_0$, take the divergence of the first equation and the time derivative of the second equation,

$$\nabla\cdot\frac{\partial \vec{u}}{\partial t}=-C^2\nabla^2 \xi$$

$$\frac{\partial^2\xi}{\partial t^2}=-\frac{\partial}{\partial t}\nabla\cdot\vec{u}$$

Finally, we substitute the first equation into the second one, exchanging the order of differentiation with respect to time and space, to arrive at a wave equation for $\xi$,

$$\frac{\partial^2\xi}{\partial t^2}=C^2\nabla^2 \xi$$

The quantity $C>0$ represents the speed of sound.

We seek a solution of this equation that is a harmonic function of time, so it oscillates with frequency $\omega$. Rather than working with sines or cosines, it is more convenient to use a complex notation, writing

$$\xi(\vec{r},t)={\rm Re}\;e^{-i\omega t}f(\vec{r})$$

The complex function $f$ satisfies the Poisson equation,

$$C^2\nabla^2 f=-\omega^2 f$$

Let's seek a solution with cylindrical symmetry, so $f(R)$ depends only on the radial coordinate $R=\sqrt{x^2+y^2}$. The Poisson equation in cylindrical coordinates takes the form

$$\frac{d^2}{dR^2}f(R)+\frac{1}{R}\frac{d}{dR}f(R)=-(\omega/C)^2f$$

The solution is a Bessel function

$$f(R)={\rm constant}\times J_0(\omega R/C)$$

The full solution thus becomes

$$\delta\rho/\rho_0=A\cos(\omega t+B)J_0(\omega R/C)$$

where $A$ and $B$ are arbitrary coeffients.

And we're done :)

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  • $\begingroup$ Thank you so much! I am still studying your solution and so far it looks great! I will be sure to finish it soon. $\endgroup$
    – user34091
    Commented May 19, 2013 at 14:42
  • $\begingroup$ I am still working to check your derivation. I am explicitly stating most of the steps you made like: $\xi=\frac{\delta\rho}{\rho_0}$ rearranges to $delta\rho=\rho_0\xi$. And substitutes into $p=p_0+C^2\delta\rho$ : $p=p_0+C^2\rho_0\xi$ I will confirm your answer today. $\endgroup$
    – user34091
    Commented May 22, 2013 at 10:20

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